Question

The equation of that diameter of the circle $x^2+y^2-6x+2y-8=0$ which passes through the origin, is

• A. $x+3y=0$
• B. $3x-y=0$
• C. $6x-y=0$
• D. $3x+2y=0$

Answer 1

The correct option is A $x+3y=0$

Given equation of a circle is
$x^2+y^2-6x+2y-8=0$
$\therefore$ Centre $=(3,-1)$
Now, the diameter also passes through the origin $O(0,0)$.
So, the required equation of the diameter is
$y-0=\frac{-1-0}{3-0}(x-0)$
$\Rightarrow y=-\frac{1}{3}(x-0)$
$\Rightarrow 3y=-x$ $\Rightarrow$ $x+3y=0$.